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Let $f : M \rightarrow N$ where $M$ and $N$ are manifolds of dimension $n$ and $m$

I don't know how to prove properly that $f$ can be a diffeomorphism only if $m=n$.

I wanted to prove that $f$ can be bijective only if $m=n$ but that's not true. So I guess the "smooth" caracter has to play something here.

StarBucK
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    See this question – lulu Apr 03 '17 at 10:06
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    This is not true. $\mathbb{R}^m$ and $\mathbb{R}^n$ have the same cardinality and so there does indeed exist a bijection between these two sets. – Oiler Apr 03 '17 at 10:07
  • You should have more constraints about $f$, otherwise your proposition is not true as @Oiler has provided an argument. – Nick Apr 03 '17 at 10:11
  • There is a setting in which $\mathbb{R}^n$ and $\mathbb{R}^m$ are different. They are different as topological spaces: there is no homeomorphism from one space to another. The obstruction to this homeomorphism is that these topological spaces have different topological dimension. – Evgeny Apr 03 '17 at 10:20
  • I edited my question to be closer to my problem. – StarBucK Apr 03 '17 at 10:22

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This is a consequnce of the famous from https://en.wikipedia.org/wiki/Invariance_of_domain

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    You don't need something that complicated in the smooth setting. Diffeomorphisms induce isomorphisms on tangent spaces, from which the result follows immediately. – user2520938 Apr 03 '17 at 10:35